i need help with this one
thanks
A particle starts from rest and is acted on by a net force that does work at a rate that is proportional to the time t. The speed of the particle is proportional to:
a. sq. root t
b. t
c. t^2
d.1/sq. root t
e. 1/t
Work done is proportional to the integral of a constant times t. That means the work done is proportional to t^2. It will equal the final kinetic energy, which is proportional to v^2. Therefore v^2 is proportional to t^2, and v must be proportional to t.
the answer is A
work is proportional to t.
work is the change in kinetic energy.
then work is of the order of V^2.
then V^2 is prop. to t.
therefore, V is prop. to sq.root t
Draw a sketch to show the force acting on the climber
the sum of 3 and C
Well, in simpler terms, the speed of the particle is like a relationship status on Facebook - it's complicated! But fear not, my friend, I'm here to clear things up.
Now, we know that the net force is doing work at a rate that is proportional to time, which means the work done is increasing as time goes on. And what happens when more work is done? That's right, the particle's speed increases!
But how does the speed relate to time? Well, let's think about it. If the work done is proportional to t^2, and the work done is equal to the final kinetic energy, which is proportional to v^2, then v^2 must also be proportional to t^2.
Now, what does that mean for the speed itself? Well, if v^2 is proportional to t^2, then that means v must be proportional to... (drumroll please)... t! So, the speed of the particle is proportional to time.
Therefore, the answer is (b) t. The speed of the particle is like a dedicated friend who sticks with you through thick and thin - it grows alongside time. So, keep on ticking, and the speed will keep on picking up!
To find the relationship between the speed of the particle and time, we can follow these steps:
1. Given that the net force doing work on the particle is proportional to time, we have the equation: F = k * t, where F is the force and k is the proportionality constant.
2. Work done is defined as the integral of force over distance: W = ∫ F * dx.
3. Since the particle starts from rest, we know that the total work done is equal to the final kinetic energy of the particle: W = (1/2) * m * v^2, where m is the mass of the particle and v is its speed.
4. We need to find the relationship between v and t, so let's set up the equation: (1/2) * m * v^2 = k * t^2.
5. Rearranging the equation, we get: v^2 = 2 * (k/m) * t^2.
6. Since v^2 is proportional to t^2, we can conclude that the speed of the particle is proportional to the square root of t.
Therefore, the answer is (a) square root t.